Question

1. $9m + 18 = \frac{1}{7} - \frac{4m + 36}{7m}$

Explanation:

Step1: Eliminate denominators

Multiply each term by \(7m\) (the least common denominator) to get rid of the fractions:
\(7m(9m + 18)=7m\times\frac{1}{7}-7m\times\frac{4m + 36}{7m}\)
Simplify each term:
\(63m^2 + 126m=m-(4m + 36)\)

Step2: Simplify the equation

Expand and combine like terms:
\(63m^2 + 126m=m - 4m - 36\)
\(63m^2 + 126m=-3m - 36\)

Step3: Rearrange into standard quadratic form

Bring all terms to the left side:
\(63m^2 + 126m + 3m + 36 = 0\)
\(63m^2 + 129m + 36 = 0\)

Step4: Simplify the quadratic equation

Divide all terms by 3:
\(21m^2 + 43m + 12 = 0\)

Step5: Solve the quadratic equation

Use the quadratic formula \(m=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\), where \(a = 21\), \(b = 43\), \(c = 12\):
First, calculate the discriminant \(\Delta=b^2 - 4ac=(43)^2 - 4(21)(12)=1849 - 1008 = 841\)
Then, \(m=\frac{-43\pm\sqrt{841}}{42}=\frac{-43\pm29}{42}\)

Step6: Find the two solutions

For the plus sign: \(m=\frac{-43 + 29}{42}=\frac{-14}{42}=-\frac{1}{3}\)
For the minus sign: \(m=\frac{-43 - 29}{42}=\frac{-72}{42}=-\frac{12}{7}\)

Step7: Check for extraneous solutions

Substitute \(m = -\frac{1}{3}\) and \(m = -\frac{12}{7}\) back into the original equation to ensure \(7m
eq0\) (denominator cannot be zero). Both values satisfy \(7m
eq0\), so they are valid.

Answer:

\(m = -\frac{1}{3}\) or \(m = -\frac{12}{7}\)